Problem 14 - Physics

Question

Location \(A\) is \(3.00 \mathrm{~m}\) to the right of a point charge \(q .\) Location \(B\) lies on the same line and is \(4.00 \mathrm{~m}\) to the right of the charge. The potential difference between the two locations is \(V_{B}-V_{A}=45.0 \mathrm{~V}\). What is the magnitude and sign of the charge?

Step-by-Step Solution

Answer

The charge is \(-6.10 \times 10^{-10} \, \mathrm{C}\). It is negative.

Step 1: Understand the Problem

We are given two points, A and B, placed at distances of 3.00 m and 4.00 m from a point charge q respectively, and the potential difference between these two points is \(V_{B} - V_{A} = 45.0\, \mathrm{V}\). We need to find the magnitude and sign of charge q.

Step 2: Recall the Potential Due to a Point Charge

The electric potential due to a point charge at a distance \(r\) from the charge is given by \(V = \frac{kq}{r}\), where \(k = 8.99 \times 10^9 \, \mathrm{N}\,\mathrm{m}^2/\mathrm{C}^2\) is the Coulomb's constant and \(q\) is the charge.

Step 3: Write the Expression for Potential Difference

The potential difference \(V_{B} - V_{A}\) can be expressed using the potentials at points B and A: \[V_{B} - V_{A} = \frac{kq}{r_B} - \frac{kq}{r_A}\] where \(r_B = 4.00\, \mathrm{m}\) and \(r_A = 3.00\, \mathrm{m}\).

Step 4: Substitute Known Values

Substitute \(V_{B} - V_{A} = 45.0 \, \mathrm{V}\), \(r_B = 4.00\, \mathrm{m}\), and \(r_A = 3.00\, \mathrm{m}\) into the potential difference formula:\[45.0 = kq \left(\frac{1}{4.00} - \frac{1}{3.00}\right)\]

Step 5: Simplify the Expression

Calculate \(\frac{1}{4.00} - \frac{1}{3.00}\):\[\frac{1}{4.00} = 0.25, \quad \frac{1}{3.00} \approx 0.333\]Thus, \[\frac{1}{4.00} - \frac{1}{3.00} = 0.25 - 0.333 = -0.083\]

Step 6: Solve for the Charge q

Plug the simplified expression into the equation:\[45.0 = kq (-0.083)\]Rearrange to solve for \(q\):\[q = \frac{45.0}{k(-0.083)}\]Calculate using \(k = 8.99 \times 10^9 \, \mathrm{N}\,\mathrm{m}^2/\mathrm{C}^2\):\[q = \frac{45.0}{(-0.083) \times 8.99 \times 10^9}\]\[q \approx -6.10 \times 10^{-10} \, \mathrm{C}\]

Step 7: Interpret the Result

The negative sign indicates that the charge is negative, and the magnitude of the charge is \(6.10 \times 10^{-10} \, \mathrm{C}\).

Key Concepts

Electric Potential Difference

Electric potential difference is an essential concept in electrostatics. It describes the work done in moving a unit positive charge from one point to another in an electric field. Essentially, it is the difference in electric potential energy between two points per unit charge.

In our exercise, points A and B have different electric potentials due to their varying distances from the point charge q. The potential difference between these points, given as \(V_B - V_A = 45.0 \, \text{V}\), tells us how much electric potential energy difference is present per unit charge between these two points.

Electric potential difference is critical when analyzing the energy changes in systems involving electric fields. It's measured in volts \( (\text{V}) \), where 1 volt is equal to 1 joule per coulomb \((1 \, \text{V} = 1 \, \text{J/C})\). This measurement helps us understand the capability of an electric field to do work.

Coulomb's Law

Coulomb's Law is a fundamental principle used to describe the force between two charges. It states that the force is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance between them.

The formula is given as:

• \( F = k \frac{|q_1 q_2|}{r^2} \)

where:

• \(F\) is the force between the charges.
• \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N}\,\text{m}^2/\text{C}^2\).
• \(q_1\) and \(q_2\) are the charges.
• \(r\) is the distance between the charges.

Coulomb's Law applies specifically to point charges and helps predict the magnitude and direction of the force experienced by each charge due to the other. Although the exercise focuses on potential difference, understanding Coulomb's Law is crucial as it underpins the calculation of forces and potential energies in electrostatic scenarios.

Point Charge

A point charge is an idealized model used in electrostatics, where the charge is considered to be concentrated at a single point in space. This concept simplifies the analysis of electric fields and potentials since it allows for easier calculations.

In our exercise, the point charge affects the electric potential at specific locations A and B. The potential due to a point charge is calculated using the formula:

• \( V = \frac{kq}{r} \)

where:

• \(V\) is the electric potential at a distance \(r\) from the charge.
• \(k\) is Coulomb's constant.
• \(q\) is the charge of the point.
• \(r\) is the radial distance from the charge to the point of interest.

The point charge model is beneficial for theoretical studies of electrostatics. For practical applications with extended charge distributions, other models and more complex calculations may be needed. However, the point charge provides a foundational understanding of how charges interact with their surroundings through electric fields and potential differences.